1. **Problem statement:** Given the joint density function $$f(x,y) = k(2x + y), \quad 2 < x < 6,\ 0 < y < 5$$ and zero otherwise, find (c) $P(X > 2.5)$. 2. **Recall:** Probability that $X > 2.5$ is found by integrating the joint density over $x > 2.5$ and the entire range of $y$: $$P(X > 2.5) = \int_{2.5}^{6} \int_{0}^{5} f(x,y) \, dy \, dx.$$ First, we need $k$ from part (a) to normalize the density. 3. **Find $k$:** From (a), the total integral over $2 < x < 6$ and $0 < y < 5$ equals 1: $$\int_{2}^{6} \int_0^{5} k(2x + y) \, dy \, dx = 1.$$ Integrate w.r.t $y$: $$\int_0^{5} (2x + y) \, dy = 2x\cdot 5 + \frac{5^2}{2} = 10x + 12.5.$$ Thus, $$k \int_2^6 (10x + 12.5) dx = 1.$$ Integrate w.r.t $x$: $$\int_2^6 (10x + 12.5) dx = 5x^2 + 12.5x \Big|_2^6 = (5 \times 36 + 75) - (5 \times 4 + 25) = (180 + 75) - (20 + 25) = 255 - 45 = 210.$$ So, $$k \times 210 = 1 \implies k = \frac{1}{210}.$$ 4. **Calculate $P(X > 2.5)$:** $$P(X > 2.5) = \int_{2.5}^6 \int_0^5 \frac{1}{210}(2x + y) \, dy \, dx.$$ Integrate w.r.t $y$: $$\int_0^5 (2x + y) dy = 10x + 12.5$$ So, $$P(X > 2.5) = \frac{1}{210} \int_{2.5}^6 (10x + 12.5) dx.$$ Integrate w.r.t $x$: $$\int_{2.5}^6 (10x + 12.5) dx = 5x^2 + 12.5x \Big|_{2.5}^6 = (5 \times 36 + 75) - (5 \times 6.25 + 31.25) = (180 + 75) - (31.25 + 31.25) = 255 - 62.5 = 192.5.$$ Finally, $$P(X > 2.5) = \frac{192.5}{210} = \frac{385}{420} = \frac{11}{12} \approx 0.9167.$$ **Answer:** $$P(X > 2.5) = \frac{11}{12} \approx 0.9167.$$